Find $\lim_{x\to -5}\dfrac{(x+1)^2}{8-x}$. Choose 1 answer: Choose 1 answer: (Choice A) A $\dfrac{16}{13}$ (Choice B) B $2$ (Choice C) C $12$ (Choice D) D The limit doesn't exist
Explanation: Let's try to find the limit using direct substitution. $\begin{aligned} \lim_{x\to -5}\dfrac{(x+1)^2}{8-x}&=\dfrac{((-5)+1)^2}{8-(-5)} \\\\ &=\dfrac{(-4)^2}{13} \\\\ &=\dfrac{16}{13} \end{aligned}$ We got a finite number. Since $\dfrac{(x+1)^2}{8-x}$ is continuous across its domain, we can determine that $\lim_{x\to -5}\dfrac{(x+1)^2}{8-x}$ is indeed equal to $\dfrac{16}{13}$. In conclusion, $\lim_{x\to -5}\dfrac{(x+1)^2}{8-x}=\dfrac{16}{13}$.